Explanation

(b)
We know,$\frac{t_{1}}{t_{2}} = \left\lbrack \frac{a_{2}}{a_{1}} \right\rbrack^{n - 1}$
Where, $t_{1}$ and $t_{2}$ are the half-life periods when the initial concentration are $a_{1}$and $a_{2}$
According to question
$t_{1} = 0.1s,a_{1} = 200$
$t_{2} = 0.4s,a_{2} = 50$
On substitution the values
$\frac{0.1}{0.4} = \left\lbrack \frac{50}{200} \right\rbrack^{(n - 1)}$
On taking log on both sides
$\log\frac{0.1}{0.4} = (n - 1)\log\frac{50}{200}$
$0.60 = (n - 1) \times 0.60$
$(n - 1) = \frac{0.60}{0.60} = 1$
$n - 1 = 1$
$n = 1 + 1 = 2$